Fourier Analysis and Number theory I: The self-dual function e^{-\pi x^2}

During a lecture at UCLA, Serge Lang asked what is the most important function in mathematics. This question is quite personal and every person certainly has his own opinion. In fact, a professor spoke out loud that the constant function is the most important function.

Serge Lang suggested {f(x) = e^{-\pi x^2}}. The function is self-dual (i.e. its Fourier transform is itself, see below) and the property is essential to the functional equation of the Riemann zeta function.

For {f \in L^1({\mathbb R})}, the Fourier transformof {f} is a function {\hat{f}} on {{\mathbb R}} defined by

\displaystyle \widehat{f}(t) = \int_{{\mathbb R}} f(x)e^{-2\pi i x t} dx.

Since {f \in L^1({\mathbb R})}, the above integral is absolutely convergent.

Theorem 1 The function {f(x) = e^{-\pi x^2}} is self-dual, i.e., {\hat{f}(t) = f(t)}.

Proof 1.We will use Cauchy’s theorem from the theory of complex analysis. For another proof without the theory of complex analysis, see proof 2 below.

\displaystyle \widehat{f}(t) = \int_{-\infty}^\infty e^{-\pi x^2-2\pi i x t} dx = \int_{-\infty}^\infty e^{-\pi (x+i t)^2} e^{-\pi t^2} dx = e^{-\pi t^2} \int_{\Im z= t} e^{-\pi z^2} dz.

We will show that the integral on the right hand side is {1} and completes the proof. Let {\Gamma_T} be the contour of the rectangle with vertices {-T}, {T}, {T+it}, {-T+it}. By Cauchy’s theorem

\displaystyle 0 = \int_{\Gamma_T} e^{-\pi z^2} dz

\displaystyle =\int_{-T}^T e^{-\pi x^2} dx + \int_0^t e^{-\pi (T+yi)^2} i dy - \int_{-T}^T e^{-\pi (x+it)^2} dx - \int_0^t e^{-\pi (-T+iy)^2} idy.

Let {T\rightarrow \infty}, the second term and the fourth term {\rightarrow 0}. Thus

\displaystyle \int_{\Im z = t} e^{-\pi z^2} dz = \int_{{\mathbb R}} e^{-\pi x^2} dx = 1.

The last equality is a standard application of double integration in polar coordinates. {\square}
 Before we proceed to the second proof, we need the following lemma.

Lemma 2 Suppose {f \in L^1}.
(i)If {f} has a continuous derivative and {f' \in L^1({\mathbb R})}, then

\displaystyle \widehat{f'}(t) = 2\pi i t \widehat{f}(t).

(ii) Let {(Mf)(x) = xf(x)}. Suppose {(Mf)(x)} is integrable, then

\displaystyle \widehat{Mf}(t) = -\frac 1{2\pi i} \frac{d}{dt} \widehat{f}(t).

Proof. Suppose {f' \in L^1({\mathbb R})}

\displaystyle \widehat{f'}(t) = \int_{-\infty}^\infty f'(x) e^{-2\pi i x t} dx

\displaystyle = \left. f(x) e^{-2\pi i x t} \right|_{-\infty}^{\infty} + 2 \pi i t \int_{-\infty}^\infty f(x) e^{-2\pi i x t} dx = 2 \pi i t \hat{f}(t).

In the last step we use the fact that {f \in L^1({\mathbb R})} and hence {f(\pm \infty) = 0}. This proves (i). Next Suppose {xf(x)} is integrable. Then {\frac{d}{dt} f(x) e^{-2\pi i x t} = -2\pi i x f(x) e^{-2\pi i x t}} is also integrable. Hence

\displaystyle \frac{d}{dt} \widehat{f}(t) = \frac{d}{dt} \int_{\mathbb R} f(x) e^{-2\pi i x t} dx

\displaystyle = -2\pi i \int_{\mathbb R} x f(x) e^{-2\pi i x t} dx = -2\pi i \widehat{Mf}(t).

Thus (ii) follows. {\square}

Proof 2. Let {f(x) = e^{-\pi x^2}}. Simple calculation shows that

\displaystyle f'(x) + 2 \pi x f(x) = 0.

Applying the Fourier transform to the above equation, by the above lemme we obtain

\displaystyle 2 \pi i t \widehat{f}(t) + i \frac{d}{dt} \widehat{f}(t) = 0.

Solving the ordinary differential equation, we obtain

\displaystyle \widehat{f}(t) = C e^{-\pi t^2},

where {C} is a constant. To find {C}, we set {t=0}

\displaystyle C = \widehat{f}(0) = \int_{{\mathbb R}} e^{-\pi x^2} dx = 1.

{\square}

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